Explanation Given: ∫ ∫ ∫ B ( x y + z ) d V , B = { 0 ≤ x ≤ 2 , 0 ≤ y ≤ 1 , 1 ≤ z ≤ 3 } Calculation: Consider the integral ∫ ∫ ∫ B ( x y + z ) d V Putting the values of the boundary conditions, the integral can be rewritten as ∫ ∫ ∫ B ( x y + z ) d V = ∫ 0 2 ∫ 0 1 ∫ 1 3 ( x y + z ) d z d y d x = ∫ 0 2 ∫ 0 1 [ x y z + 1 2 z 2 ] z = 1 z = 3 d y d x = ∫ 0 2 ∫ 0 1 [ 2 x y + 4 ] d y d x = ∫ 0 2 [ x y 2 + 4 y ] y = 0 1 d x = ∫ 0 2 [ x + 4 ] d x = [ x 2 2 + 4 x ] 0 2 = [ 1 2 ( 4 ) + 4 ( 2 ) ]

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 16, Problem 27CRE

To determine

To calculate:

$\int \int \int_{B} (x y + z) d V$ where $B = {0 \leq x \leq 2, 0 \leq y \leq 1, 1 \leq z \leq 3}$ , as an iterated integral in two ways.

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Chapter 16, Problem 26CRE

Chapter 16, Problem 28CRE

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Chapter 16 Solutions

CALCULUS (CLOTH)

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∫ ∫ ∫ B ( x y + z ) d V where B = { 0 ≤ x ≤ 2 , 0 ≤ y ≤ 1 , 1 ≤ z ≤ 3 } , as an iterated integral in two ways.

Solution Summary: The author explains how the integral can be rewritten as x y + z d V, where the values of the boundary conditions are given.

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To calculate:

$\int \int \int_{B} (x y + z) d V$ where $B = {0 \leq x \leq 2, 0 \leq y \leq 1, 1 \leq z \leq 3}$ , as an iterated integral in two ways.

Want to see the full answer?

Chapter 16 Solutions

∫ ∫ ∫ B ( x y + z ) d V where B = { 0 ≤ x ≤ 2 , 0 ≤ y ≤ 1 , 1 ≤ z ≤ 3 } , as an iterated integral in two ways.

Solution Summary: The author explains how the integral can be rewritten as x y + z d V, where the values of the boundary conditions are given.

Videos

To calculate: ∫∫∫B(xy+z)dVwhere B={0≤x≤2,0≤y≤1,1≤z≤3}, as an iterated integral in two ways.

Want to see the full answer?

Chapter 16 Solutions

To calculate:

$\int \int \int_{B} (x y + z) d V$ where $B = {0 \leq x \leq 2, 0 \leq y \leq 1, 1 \leq z \leq 3}$ , as an iterated integral in two ways.